Civil Engineering - Hydraulics - Discussion

Given, Atmospheric pressure P, oil = 0.1kg/cm² = 10000N/m².

{How I converted 0.1kg/cm² to N/m².
For this, we should know.
1kg = 10N & 1m = 100Cm.
So, (0.1x10)N÷(100-¹x100-¹)m².
= 10000N/m²}.

Density of oil = Specific gravity of oil x Density of water.
=>0.8x1000 = 800kg/m³.

The pressure at 2.5 m depth of oil surface will be:

P = Poil+{Density of oil (p)x Acc due to gravity(g) x depth of oil surface(h)}.
P = 10000N/m²+(800x10x2.5)N/m²,
P = 30000N/m² or Pa.

Now, converting to water head:

P = {Density of water(p) x Acc due to gravity(g) x depth of water(h)}
=>30000N/m²=1000x10xh
h = 3 meters of the water column.

Sure, let me explain in more detail.

The pressure exerted by a fluid increases with depth. In this case, we have an oil tank with a specific gravity of 0.8 and an atmospheric pressure of 0.1 kg/cm2 on the surface. To find the pressure at a depth of 2.5 m, we can use the formula:

Pressure = Atmospheric pressure + (Density of fluid x Acceleration due to gravity x Depth).

Since the specific gravity of oil is less than 1, we can use the formula:
Pressure = Atmospheric pressure + (Specific gravity x Atmospheric pressure x Depth).

Plugging in the values, we get:
Pressure = 0.1 kg/cm2 + (0.8 x 0.1 kg/cm2 x 2.5 m)
Simplifying the equation, we find that the pressure at a depth of 2.5 m is equivalent to the pressure exerted by 3 m of water.

0.1kg/cm^2 = 1m,

Pressure of 5m water = pressure of 5m* sp.gravity,
Or, = 2.5 * 0.8,= 2

Therefore
(1+2)= 3m.

Given, S = 0.8, Patm = 0.1 kg/cm2 = 10000 N/m2.
Density of Oil= S x ρ(w)= 0.8x1000= 800 kg/m3.
Pressure at 5m below surface.

P = Patm+ ρ(oil)*g*h = 10000 + (800 * 10 * 2.5)= 30000 N/m2.
Now for water head.
P=ρ(w)*g*h.
30000=100*10*h,
h = 3m of water.

P(total pressure)= pressure of oil at 2.5m + p(ATM)
= 0.8X1000X9.81X2.5 + 0.1X9.81X10^4 (kg is converted into Newton by multiplying 9.81 & cm^2 is also converted into m^2).
= 29430 N/m^2.

Now,
The total pressure = pressure term of water.
29430 = 1000X9.81XH,
or, H = 29430/9810,
H = 3 m.
Hence, option C. 3 m of water is correct.

If the atmospheric pressure on the surface of an oil tank (sp. gr. 0.8) is 0.1 kg/cm2, the pressure at a depth of 2.5 m, is 3m.
P-atm = 0.1kg/cm2=0.1*10000=1000kg/m2=1000*9.81=9810 N/m2=9810/1000=9.81 KN/m2.
P-oil => p=0.8*1000 of watr=800kg/m3.

As P=pgh = 800*9.81*2.5=19620 N/m2=196209/1000 = 19.62 KN/m2.

P-total = P-atm + P-oil.
P-total = 19.62 + 9.81 = 29.43 KN/m2.
P in-term of water =>pg of water = 9.81KN/m3 so 29.43/9.81 = 3m height of water.

P= atmospheric pressure + ρ*g*h.

P= 0.1 kg/cm2 + 0.8*1000*10*2.5=10000N/m2 +20000N/m2,
= 30000N/ m2.

Now convert into water head as,
30000= 1000*10* height of water.
Height of water = 3.

Pressure at 2.5m depth due to gravity = 2.5 * 0.8 = 2m of WC.
Pressure at surface = 0.1 kg/cm^2 = 1 m of WC (as 10 m WC = 1 kg/cm^2),
Hence, Total pressure at 2.5m depth = (2 + 1) m of WC = 3 m of WC.

If we will consider the value of g then the pressure will come in the unit of Newton/square meter.

For determining the value of pressure in the form of kg/m2 we can't use Value of g.

How the pressure at 250 cm comes out as 2000 Kg/m2?

That will be $gh, we don't consider 'g' value.

Can anyone explain me?